Solve for $x$ and $y$ using elimination. ${6x-y = 12}$ ${5x+4y = 39}$
Answer: We can eliminate $y$ by adding the equations together when the $y$ coefficients have opposite signs. Multiply the top equation by $4$ ${24x-4y = 48}$ $5x+4y = 39$ Add the top and bottom equations together. $29x = 87$ $\dfrac{29x}{{29}} = \dfrac{87}{{29}}$ ${x = 3}$ Now that you know ${x = 3}$ , plug it back into $\thinspace {6x-y = 12}\thinspace$ to find $y$ ${6}{(3)}{ - y = 12}$ $18-y = 12$ $18{-18} - y = 12{-18}$ $-y = -6$ $\dfrac{-y}{{-1}} = \dfrac{-6}{{-1}}$ ${y = 6}$ You can also plug ${x = 3}$ into $\thinspace {5x+4y = 39}\thinspace$ and get the same answer for $y$ : ${5}{(3)}{ + 4y = 39}$ ${y = 6}$